Proof. Set $v=v(E)$ for temporary notation. Firstly, for any $G\subseteq F \subseteq E$ in $\operatorname {\mathrm {Coh}}^{\beta (\sigma )}(X)$ satisfying that $E,F,G$ have the same $\sigma $ -phase, we always have that $v(F/G)$ is lying in the parallelogram $ {\mathbf {P}}_{Ov^{\!+}_{\sigma } v v^{\!-}_{\sigma }}$ . This is because there are inclusions

$$ \begin{align*}{\mathrm{L}}_{[0, Z_{\tau}(F/G) ]} \subseteq {\mathrm{L}}_{[ 0, Z_{\tau}(F) ] } \subseteq {\mathrm{L}}_{[0, Z_{\tau}(E) ]}\end{align*} $$

for any $\tau \in {\mathrm {L}}_{(\sigma ',\sigma " )}$ , which yields that

$$ \begin{align*} v(F/G) \in \bigcap_{\tau \in {\mathrm{L}}_{( \sigma', \sigma" )}} Z_\tau^{-1}({\mathrm{L}}_{[ 0,Z_\tau(v)]})={\mathbf{P}}_{Ov^+_{\sigma} v v^-_{\sigma}}. \end{align*} $$

In particular, if $0=\widetilde {E}_0 \subset \dots \subset \widetilde {E}_k=E$ is a $\sigma $ -Jordan–Holder filtration of E with $E_i = \widetilde {E}_i / \widetilde {E}_{i-1}$ its JH-factors, then any $v(E_i)$ and also $v-v(E_i)$ is lying ${\mathbf {P}}_{Ov^+_{\sigma } v v^-_{\sigma }}$ .

If necessary, reordering these factors $E_i$ such that the angles between ${\mathrm {L}}^+_{O,v(E_i)}$ and ${\mathrm {L}}_{O,v_\sigma ^+}^+$ increase with respect to i. And we get a polygon with vertexes $\sum _{i=1}^j v(E_i)$ ( $0\leq j\leq k$ ). Since $v(E_1)$ and $v-v(E_k)$ lie in $\mathbf {P}^+_{Ov^+_\sigma v}$ , one sees that $\sum \limits _{i=1}^{j} v(E_i)$ is an integer point in ${\mathbf {P}}_{O v_\sigma ^+ v}\setminus {\mathrm {L}}_{(O,v)} $ for any j. We claim that either $v(E_1)$ or $\sum \limits _{i=1}^{k-1} v(E_i)$ is lying in $\Omega _v(\sigma )$ . This can be proved by using purely Euclidean geometry. Suppose this fails, then we have

(3.2) $$ \begin{align} \begin{aligned} v(E_1)^2\geq -2 \quad \mbox{and}\quad (v(E_1)-v)^2 <-2 , \\ (\sum\limits_{i=1}^{k-1} v(E_i) -v)^2\geq -2 \quad \mbox{and}\quad (\sum\limits_{i=1}^{k-1} v(E_i) )^2<-2, \end{aligned} \end{align} $$

as $E_i$ is stable. If we restrict the quadratic equation $u^2=-2$ to the plane of ${\mathbf {P}}_{ovv_\sigma ^+}$ , we can obtain a hyperbola whose center is O. The edge ${\mathrm {L}}_{[O,\, v_\sigma ^+]}$ can meet the connected component of this hyperbola at most one point. Similarly, ${\mathrm {L}}_{[v,\, v_\sigma ^+]} $ can intersect with the connected component of the hyperbola defined by $(u-v)^2=-2 $ at most one point. Note that the edge ${\mathrm {L}}_{[O,v]}$ is lying outside the area

(3.3) $$ \begin{align} \Big\{u^2<-2~\mbox{and}~(u-v)^2<-2\Big\}. \end{align} $$

See the shadow part in Figure 5. By Equation (3.2), $v_\sigma ^+$ has to lie in the region (3.3). Moreover, as one can see from the picture, there is a point $w\in {\mathbf {P}}_{Ov^+_\sigma v}^\circ $ lying on the intersection of two hyperboloids

(3.4) $$ \begin{align} \Big\{u\in \mathbb{R}^3|~ u^2= -2\Big\} \cap \Big\{u\in\mathbb{R}^3|~(u-v)^2=-2\Big\}, \end{align} $$

and the line ${\mathrm {L}}_{w,v^+_\sigma }$ will intersect the edge ${\mathrm {L}}_{[O,v]}$ at a point, denoted by q. Thus, we get that the point $v(E_1)$ is contained in the triangle ${\mathbf {P}}_{qv v^+_\sigma }$ , while the point $\sum _{i=1}^{k-1} v(E_i)$ is contained in the triangle $ {\mathbf {P}}_{Oq v^+_\sigma }$ . If we define ${\mathrm {L}}_u$ to be the line passing through $u\in \mathbb {R}^3$ and parallel to the line ${\mathrm {L}}_{w, v^+_{\sigma } }$ , the discussion above exactly means

(3.5) $$ \begin{align} {\mathrm{L}}_{[O,\, p_1]} \subset {\mathrm{L}}_{[O,\,q)}\subset {\mathrm{L}}_{[O,\,p_2]}, \end{align} $$

where $p_1={\mathrm {L}}_{\sum _{i=1}^{k-1} v(E_i)}\cap {\mathrm {L}}_{[O,\,v]} $ and $p_2={\mathrm {L}}_{v(E_1)}\cap {\mathrm {L}}_{[O,\,v]}$ ; see Figure 5.

Figure 5 If this plane corresponds to a wall, there must be some integer point inside this triangle and below these two hyperbolas.

Next, one can regard ${\mathrm {L}}_{w,v^+_\sigma }$ as a stability condition in a natural way. Consider the line $\ell $ passing through the origin O on the plane $\Pi $ , which is parallel to the line passing through w and $v_{\sigma }^+$ . By construction, the line $\ell $ lies between $[\sigma ']$ and $[\sigma "]$ . Thus, there is a stability condition $\tau \in (\sigma ', \sigma ") $ such that $[\tau ]$ corresponds to the line $\ell $ . Since $Z_\tau (\ell ) = 0$ , we obtain $Z_\tau (E_1) = Z_\tau (p_2)$ , and the same holds for $p_1$ .

Using the inclusions (3.5), we obtain the inequality

(3.6) $$ \begin{align} \frac{\big\lvert Z_{\tau}(v(E_1)) \big\rvert}{\big\lvert Z_{\tau}(\sum_{i=1}^{k-1} v(E_i))\big\rvert }=\frac{\big\lvert Z_{\tau}(p_2)\big\rvert }{ \big\lvert Z_{\tau}(p_1) \big\rvert } = \frac{\|{\mathrm{L}}_{[O,\,p_2]}\|}{\|{\mathrm{L}}_{[O,\,p_1]}\|}>1. \end{align} $$

However, this contradicts to the relation $ \sum _{i=1}^{k-1}\big \lvert Z_{\tau }(v(E_i)) \big \rvert = \big |Z_{\tau } \left (\sum _{i=1}^{k-1} v(E_i) \right )\big |$ which finishes the proof.

Proof. The ‘if’ part follows directly from Lemma 3.2. For the ‘only if’ part, suppose there exists some stability condition $\sigma $ and an integer point $w \in \Omega _v(\sigma )$ . Then, we can find $\sigma $ -stable objects $F_1$ and $F_2$ such that $v(F_1)=w$ and $v(F_2)=v-w$ , and $\sigma $ will be lying on a wall of from the construction. For the strictly semistable case, one just notes that the Mukai vectors of all the factors are lying on ${\mathrm {L}}_{(O,v)}$ .

Proof. (i). By Proposition 3.1, it will be sufficient to show that

$$ \begin{align*}\mathbf{P}^\circ_{o \pi_v \infty} \subseteq V(X).\end{align*} $$

Due to the explicit description of $V(X)$ in (2.1), this is equivalent to showing that there is no projection of root lying in $\mathbf {P}^\circ _{o\pi _v\infty }$ . Suppose there exists a root $\delta =(r',c',s')\in \mathrm {R}(X)$ with $\pi _\delta \in \mathbf {P}^\circ _{o\pi _v\infty }$ . Then we have

(4.3) $$ \begin{align} \frac{c'}{r' }<\frac{c}{r}\quad\text{and}\quad\frac{c'}{s'}<\frac{c}{s}, \end{align} $$

see Figure 7. Note that $2rs=c^2(2g-2)$ and $2r's'=(c')^2(2g-2) +2$ , one can plug into (4.3) to get

(4.4) $$ \begin{align} \frac{r}{\gcd(r,c)} c' < \frac{c}{\gcd(r,c)} r'< \frac{r}{\gcd(r,c)} c' + \frac{r}{\gcd(r,c)(g-1)c'}\leq \frac{r}{\gcd(r,c)} c'+1. \end{align} $$

which is not possible.

(ii). According to Proposition 3.4, we just need to show that $ \Omega ^+_{v}(\sigma )\cap {\mathrm {H}}_{\text {alg}}^{*}(X) =\emptyset $ for any $\sigma \in {\mathbf {P}}_{o\pi _v \infty } \cap \Gamma $ . Suppose there is an integer point $(x,y,z)\in \Omega ^+_v( \sigma _0) $ for some $\sigma _0\in {\mathbf {P}}_{o\pi _v \infty } $ . By the construction of $\Omega ^+_{v}({\mathbf {P}}_{o\pi _v \infty } \cap \Gamma )$ , we have $0<y\leq c$ and the point $(x,y,z)$ is lying in the interior of the triangle ${\mathbf {P}}_{u_1 u_2 u_3}$ with vertices

$$ \begin{align*}u_1=(\frac{ry}{c}, y,\frac{gcy}{r}), u_2=(\frac{ry}{c}, y,\frac{sy}{c}) \text{ and } u_3=(\frac{gcy}{s}, y,\frac{sy}{c}). \end{align*} $$

As $ y^2(g-1)+1\geq xz$ and $z\geq \frac {sy}{c}$ , one has

$$ \begin{align*}\frac{ry}{c}<x< \frac{y^2(g-1)+1}{sy/c}.\end{align*} $$

Note that $c^2(g-1)-\frac {v^2}{2}=rs$ , the condition $s=\lfloor \frac {(g-1)c^2+1}{r} \rfloor $ is equivalent to $r>\frac {v^2}{2}+1$ . Then we have

(4.5) $$ \begin{align} \begin{aligned} 0< \frac{y^2(g-1)+1}{sy/c}- \frac{ry}{c} &< \max\left\{\frac{gc}{s}-\frac{r}{c}, \frac{c^2(g-1)+1}{s}-r\right\} \\ & = \max\left\{\frac{r(c^2+\frac{v^2}{2})}{c(c^2(g-1)-\frac{v^2}{2})}, \frac{r(\frac{v^2}{2}+1)}{c^2(g-1)-\frac{v^2}{2}}\right\}\\ &\leq \max\left\{\frac{r(c^2+r-2)}{c(c^2(g-1)-r+2)}, \frac{r^2-r}{c^2(g-1)-r+2}\right\} \\& \leq \frac{1}{c}. \end{aligned} \end{align} $$

Here, the last inequality follows from our assumption $g-1\geq \max \{\frac {r^2}{c},r+1\}$ . This means $0<x-\frac {ry}{c}<\frac {1}{c}$ which contradicts to the fact x is an integer.

Proof. The condition $mr>c>0$ ensures $\pi _v$ lies in the first quadrant while $\pi _{v(-m)}$ lies in the second quadrant, and the condition $s(mr-c)>rc$ ensures $\sigma _v$ is below $o'$ . The assertion then follows from the direct computation that ${\mathrm {L}}_{(o,\sigma _v]} \subseteq {\pmb {\triangle }}_{\pi _w}({\mathrm {L}}_{[o, o']}) \cap \Gamma \subseteq {\mathbf {P}}_{o\pi _w \infty } \cap \Gamma $ for $w = v$ or $v(-m)$ .

Proof. The argument is essentially the same as the primitive case proved in [Reference Feyzbakhsh11, Proposition 4.2]. Here, we provide the details for completeness.

We first show that for any v satisfying Equation (✶−1), if both v and $v(-m)$ admit no wall in $(o, \sigma _v ]$ , then so does

Let ${\mathcal {W}}_{i_*F}$ be the first wall, and let $\sigma _{\alpha ',0}\in {\mathcal {W}}_{i_*F} $ be a stability condition. Suppose ${\mathcal {W}}_{i_\ast F}$ is below or on the line ${\mathrm {L}}_{\pi _v, \pi _{v(-m)}}$ . Then for any destabilising sequence

(5.1) $$ \begin{align} F_1 \hookrightarrow i_*F \twoheadrightarrow F_2 \end{align} $$

in $\operatorname {\mathrm {Coh}}^{\beta =0}(X)$ such that $F_1,F_2$ are $\sigma _{\alpha ',0}$ -semistable, and

(5.2) $$ \begin{align} \phi_{\alpha,0}(F_1)> \phi_{\alpha,0}(i_* F)\quad \text{for} \quad \alpha<\alpha'. \end{align} $$

Taking the cohomology of Equation (5.1) gives a long exact sequence of sheaves

(5.3) $$ \begin{align} 0 \to \operatorname{{\mathrm{H}}} ^{-1}(F_2) \to F_1 \xrightarrow{d_0} i_* F \xrightarrow{d_1} \operatorname{{\mathrm{H}}} ^0(F_2) \to 0. \end{align} $$

Set $v(F_1)=(r',c',s')$ , then we have $r'>0$ by Equation (5.2). Let T be the maximal torsion subsheaf of $F_1$ , and we can write $v(T)=(0, \hat c, \hat s)$ for some $\hat c, \hat s\in \mathbb {Z}$ . Consider the inclusions $T \hookrightarrow F_1 \hookrightarrow i_*F$ and take the cohomology, one can get

$$ \begin{align*} 0\to \operatorname{{\mathrm{H}}}^{-1}(\operatorname{\mathrm{cok}}) \to T \to i_* F \to \operatorname{{\mathrm{H}}}^0(\operatorname{\mathrm{cok}})\to 0. \end{align*} $$

Since $\operatorname {{\mathrm {H}}}^{-1}(\operatorname {\mathrm {cok}})$ is torsion-free, it must be zero. It follows that T is a subsheaf of $i_*F$ and $\operatorname {\mathrm {rk}} (i^* T)= \frac {\hat c}{m}$ . If we let $v({\mathrm {H}}^0(F_2))=(0,c",s")$ , by restricting (5.3) to the curve C, one can get

$$ \begin{align*} \begin{aligned} r'+\frac{\hat c}{m} &=\operatorname{\mathrm{rk}}(F_1 / T) + \operatorname{\mathrm{rk}}(i^*T ) \geq \operatorname{\mathrm{rk}}(i^*F_1)\\ &\geq \operatorname{\mathrm{rk}}(i^*\ker d_1) \geq \operatorname{\mathrm{rk}}(i^*F) - \operatorname{\mathrm{rk}} (i^*\operatorname{{\mathrm{H}}}^0(F_2)) = r -\frac{c"}{m}. \end{aligned} \end{align*} $$

In other words,

(5.4) $$ \begin{align} \mu(F_1/T)-\mu({\mathrm{H}}^{-1}(F_2))=\frac{c'-\hat{c}}{r'} - \frac{c'+c"-m r}{r'} \leq m. \end{align} $$

Using Lemma 5.2 below, we can take the destabilising sequence (5.1) satisfying

(5.5) $$ \begin{align} \mu^-_H(F_1/T) \geq \frac{c}{r} \,\text{ and } \, \mu^+_H(H^{-1}(F_2)) \leq \frac{c-mr}{r}. \end{align} $$

This gives

(5.6) $$ \begin{align} \mu^-_H(F_1/T)-\mu^+_H(H^{-1}(F_2)) \geq m. \end{align} $$

Combining Equations (5.4) and (5.6), we get $mr-c"-\hat {c}= mr'$ , thus

$$\begin{align*}\mu_H(F_1/T) = \frac{c'-\hat{c}}{r'} = \frac{c'-\hat{c}}{r - \frac{c"+\hat{c}}{m} } = \frac{c}{r}, \end{align*}$$

and both $F_1/T$ and $ \operatorname {{\mathrm {H}}}^{-1}(F_2)$ are $\mu _H$ -semistable. Since $\gcd (r,c)=1$ and $i_*F$ does not contain any skyscraper sheaf, we have $\hat {c}= c" =0$ and $\hat {s}=0$ . This shows $T =0$ , and hence $v(F_1)=(r, c, s')$ . Note that by our assumption, we have $\pi _{v(F_1)}\in {\mathrm {L}}_{(o,\pi _v)}$ , which means $s \leq s'$ . If $s<s'$ ; however, as $v^2 < 2r-2$ by Equation (✶-1), this gives

$$\begin{align*}\begin{aligned} v(F_1)^2&=c^2(2g-2) -2rs'\\ &=v^2+2r(s-s')\\ &< 2r(s-s'+1)-2 \leq -2 \end{aligned} \end{align*}$$

which contradicts to the fact that $F_1$ is $\mu _H$ -stable. This forces $s=s'$ and ${\mathcal {W}}_{i_\ast \mathcal {F}}\subseteq {\mathrm {L}}_{\pi _v,\pi _{v(-m)}}$ .

In the case that ${\mathcal {W}}_{i_\ast \mathcal {F}}\subseteq {\mathrm {L}}_{\pi _v,\pi _{v(-m)}}$ , we have

$$ \begin{align*}\mu^-_H(F_1/T)-\mu^+_H(H^{-1}(F_2)) = m\end{align*} $$

and $F_1$ is a stable sheaf. Note that the map $d_0:F_1 \rightarrow i_* F$ factors through $d_0'\colon i_* (F_1|_C) \to i_*F$ and $\mu _H(i_* (F_1|_C))=\mu _H(i_\ast F)$ . Applying Theorem 4.1 to $F_1$ , we know that $i_*(F_1|_C)$ is stable as well. It follows that $d_0'$ is an isomorphism.

Proof. Denote $\sigma _1 = {\mathcal {W}}_{i_*F} \cap {\mathrm {L}}_{(o,\sigma _v]}$ . By Remark 3.3, we can take the destabilising sequence

$$ \begin{align*} F_1 \hookrightarrow i_*F \twoheadrightarrow F_2 \end{align*} $$

satisfying $v(F_1) \in \Omega ^+_{v|_C}(\sigma _1) \subseteq \Omega ^+_{v|_C} ({\mathrm {L}}_{( o, \sigma _v]})$ and $v(F_2) \in \Omega ^-_{v|_C}(\sigma _1) \subseteq \Omega ^-_{v|_C} ({\mathrm {L}}_{( o, \sigma _v]})$ . We divide the proof into three steps.

Proof. When $v^2=0$ , this is essentially proved in [Reference Feyzbakhsh11, Lemma 4.3]. Let us give a slightly different argument which also works for $v^2>0$ . Suppose the HN-filtration of $i_*F$ for $\bar {\sigma }=(\operatorname {\mathrm {Coh}}^{\beta =0}(X), \overline {Z})$ is given by

(5.11) $$ \begin{align} 0 = \widetilde{E}_0 \subset \widetilde{E}_1 \subset \dots \subset \widetilde{E}_{l-1} \subset \widetilde{E}_l = i_* F \end{align} $$

with the semistable HN-factors. To show ${\mathbf {P}}_{i_\ast F}\subseteq {\mathbf {P}}_v$ , it suffices to show that

(5.12) $$ \begin{align} \phi_{\bar{\sigma}}(v)\geq \phi_{\bar{\sigma}}({E_1}) \quad \text{and} \quad \phi_{\bar{\sigma}}({E_l}) \geq \phi_{\bar{\sigma}}({v(-m)}) \end{align} $$

(see Figure 8b) since $ {\mathbf {P}}_{i_*F}$ is convex. According to the proof of Proposition 2.2, for any object in $\operatorname {\mathrm {Coh}}^0(X) $ , the angle $\phi _{\bar {\sigma }}$ in Figure 8b is an increasing function of the angle $\theta _{o'}$ in Figure 8a. (They are actually equal in this case, as $\cot \theta _{o'} = \frac {y-1}{x}=\frac {r-s}{c}=\frac {\operatorname {{\mathrm {Re}}} \overline Z}{\operatorname {{\mathrm {Im}}} \overline Z} = \cot \phi _{\bar {\sigma }}$ ). Therefore, it is equivalent to show

(5.13) $$ \begin{align} \theta_{o'}(\pi_v)\geq \theta_{o'}(\pi_{E_1}) \quad \text{and} \quad \theta_{o'}(\pi_{E_l}) \geq \theta_{o'}(\pi_{v(-m)}) \end{align} $$

in Figure 8a.

Figure 8 The angle $\theta _{o'}$ is equal to the angle $\phi _{\bar {\sigma }}$ .

To prove Equation (5.13), consider the sequence

(5.14) $$ \begin{align} 0 \to \widetilde{E}_n \xrightarrow{f_n} i_* F \to \operatorname{\mathrm{cok}}(f_n) \to 0 \end{align} $$

for each $\widetilde {E}_n$ . Since the first wall is not below ${\mathrm {L}}_{\pi _{i_* F}, \pi _v}$ , we have $\phi _{\sigma _v}(\widetilde {E}_n)\leq \phi _{\sigma _v}(i_*F)$ . As $\phi _{\bar \sigma }(\widetilde {E}_n) \geq \phi _{\bar \sigma }(i_*F)$ , there exists some stability condition $\sigma \in {\mathrm {L}}_{(o', \sigma _v]}$ such that the objects in Equation (5.14) have the same $\sigma $ -phase. As a consequence, we have

(5.15) $$ \begin{align} \pi_{\widetilde E_i} \in \bigcup_{\sigma\in{\mathrm{L}}_{[\sigma_v, o')}} {\mathrm{L}}_{\pi_{i_*F, \sigma}}.\end{align} $$

Take $n=1$ , and set $v(E_1) = (r',c',s')$ . We claim that $\pi _{E_1} \notin {\mathbf {P}}_{o' \pi _v \sigma _v} \setminus \left \{ \pi _v \right \} $ which yields $ \theta _{o'}(\pi _{E_1}) \leq \theta _{o'}(\pi _v)$ . This can be proved by cases as follows:

1. Case (1) If $v^2 =0$ , $\pi _{E_1} \notin {\mathbf {P}}_{o' \pi _v \sigma _v}\setminus \{ \pi _v\}$ automatically holds. This is because ${\mathbf {P}}_{o' \pi _v \sigma _v}\setminus \{ \pi _v,o'\} \subseteq V(X)$ and Equation (2.2).

2. Case (2) If $v^2>0$ and $r' \leq r$ or $s' \leq s$ , as $E_1$ is $\bar {\sigma }$ -semistable, we may assume $v(E_1)^2 \geq -2$ otherwise we may replace $E_1$ by its first JH-factor. According to Step 1 in Lemma 5.2, we have

   $$ \begin{align*}v(E_1) \in \Omega_v^+({\mathrm{L}}_{(o,\,o')})\end{align*} $$
   which contradicts to the assumption $\Omega _v^+({\mathrm {L}}_{(o,\,o')})\cap {\mathrm {H}}_{\text {alg}}^{*}(X)=\emptyset $ .

3. Case (3) If $r'> r$ and $s'>s$ , we claim that $r' < r+1$ . Choose a stability condition $\sigma \in {\mathrm {L}}_{[\sigma _v, o')}$ such that $\phi _{\sigma }(i_*F) = \phi _{\sigma }(E_1)$ . Then $v(E_1)\notin \{ O, v|_c\}$ is lying in the triangle ${\mathbf {P}}_{Ov|_c (v|_c)^+_{\sigma }}$ . This means we have $0< c' < mr$ and

   $$\begin{align*}g(c')^2 - r's' \geq 0, \quad g(c'-mr)^2 -r'(s'+(g-1)m(mr-2c)) \geq 0. \end{align*}$$

   After reduction, we know that $r'<r+1$ as $g c^2 -(r+1)s \leq 0 $ and $ g (c-mr)^2 -(r+1)\tilde {s} \leq 0$ by (✶-2).

Similarly, take $n=l-1$ and use the $\bar {\sigma }$ -semistability of $\operatorname {\mathrm {cok}}(f_{l-1})=E_l$ , one can prove the second inequality of Equation (5.13).

Finally, if ${\mathbf {P}}_{i_*F} = {\mathbf {P}}_v$ , the first wall will coincide with the line ${\mathrm {L}}_{\pi _{v(-m)},\pi _v}$ and the last assertion follows from Theorem 5.1.

Proof. Suppose we have ${\mathbf {P}}_v\neq {\mathbf {P}}_{i_\ast F}$ for some $F\in {\mathbf {BN}}_C(v)$ . By Proposition 5.5 and the convexity, we have

$$ \begin{align*} r+s \leq h^0(C, F)= h^0(X,i_*F) \leq \frac{\|i_* F\|+\chi}{2} \leq \frac{\hbar+\chi}{2}, \end{align*} $$

where $ \hbar =\sqrt {(r+s-\chi )^2+4(mr-c)^2}+ \sqrt {(r+s)^2+4c^2}.$ Then we get

(5.18) $$ \begin{align} \frac{\hbar + \chi}{2} - (r+s)\geq \frac{\hbar + \chi}{2} - \frac{\|i_\ast F\|+\chi}{2} =\frac{\hbar- \|i_\ast F\|}{2}. \end{align} $$

However, note that the polygon ${\mathbf {P}}_{0z^{\prime }_1z_1^{+1}z_2'z_2}$ is convex under the assumption (i), we have

(5.19) $$ \begin{align} \begin{aligned} \hbar- \|i_\ast F\| &\geq \|z_1 -z^{\prime}_1\|-\|z^{\prime}_1- z_1^{+1}\|+ \|z_1 -z^{\prime}_2\|-\|z^{\prime}_2- z_1^{+1}\|. \end{aligned} \end{align} $$

Combined with assumption (ii), we get

(5.20) $$ \begin{align} {\hbar+ \chi} - 2(r+s) &= \sqrt{(r+s)^2+4c^2}-(r+s) \nonumber\\ & \mathrel{\phantom{=}}+ \sqrt{(r+s-\chi)^2+4(mr-c)^2}-(r+s-\chi) \nonumber \\ &< \frac{2c^2}{r+s}+\frac{2(mr-c)^2}{r+s-\chi}. \end{align} $$

(5.21) $$ \begin{align} & \kern1pt\leq \hbar- \|i_\ast F\|\ \ \qquad\qquad\qquad \end{align} $$

which contradicts Equation (5.18). This proves the assertion.

Proof. As $r>1+\frac {v^2}{2}$ , we have

$$ \begin{align*}s=\lfloor \frac{(g-1)c^2+1}{r}\rfloor \geq 4rc^2.\end{align*} $$

This gives

(5.22) $$ \begin{align} \begin{aligned} s-r&\geq 4rc^2-r\geq 3c\\ s-r-\chi & =\frac{(g-1)(mr-c)^2}{r}-\frac{v^2}{2r}-r \geq 4r(mr-c)^2-r \geq 3(mr-c), \end{aligned} \end{align} $$

as $mr-c>0$ . Moreover, one can compute that

$$ \begin{align*}\frac{2s-2r-c}{2s+2r+c} \geq \frac{8rc^2-2r-c}{8rc^2+2r+c} \geq \frac{6r-1}{10r+1}> \frac{1}{r} \geq \frac{4c^2}{s}.\end{align*} $$

It follows that

$$ \begin{align*} \|z_1- z^{\prime}_1\|-\|z^{\prime}_1 -z_1^{+1}\| &= \sqrt{(\frac{s-r}{c})^2+4g}-\sqrt{(\frac{s-r}{c}-1)^2+4g} \\ &> \frac{\frac{s-r}{c}-\frac{1}{2}}{\sqrt{(\frac{s+r}{c})^2+4(1+\frac{n}{c^2})}} \\ &> \frac{2s-2r-c}{2s+2r+c}> \frac{2c^2}{s}+\frac{2c^2}{r+s} \end{align*} $$

and $\|z_1 -z^{\prime }_2\|-\|z^{\prime }_2- z_1^{+1}\| = \sqrt {(\frac {s-r-\chi }{mr-c})^2+4g} - \sqrt {(\frac {s-r-\chi }{mr-c}-1)^2+4g} \geq 0$ . The assertion can be concluded from Theorem 5.6.

Proof. By Lemma 6.2, if there is ${\mathbf {P}}_{i_\ast F}\neq {\mathbf {P}}_v$ for some F, the polygon ${\mathbf {P}}_{v}$ will be at least $3$ -sharp. Therefore, one of the following is true:

1. (i) ${\mathbf {P}}_{i_*F}$ is contained in the polygon ${\mathbf {P}}_{0z^{\prime }_1z_1^{+3} z_2' z_2}$ .

2. (ii) $z_1^{+1}=g-k^2+k\sqrt {-1}$ is a vertex of ${\mathbf {P}}_{i_*F}$

3. (iii) $z_1^{+2}=g+1-k^2+k\sqrt {-1}$ is a vertex of ${\mathbf {P}}_{i_*F}$

We will analyse them by cases. Let us first show that case (i) is impossible if $(g,k,m)\neq (5,3,3)$ . By Equation (5.18), it suffices to show that

(6.2) $$ \begin{align} \hbar- \|i_\ast F\|>\hbar+\chi-2(g-1+k^2). \end{align} $$

when ${\mathbf {P}}_{i_*F}\subseteq {\mathbf {P}}_{0z^{\prime }_1z_1^{+3} z_2' z_2}$ . Set $\tilde {k} = m(g-1)-k$ . As in the proof of Theorem 5.6, from the convexity and a direct computation, one can get

(6.3) $$ \begin{align} \begin{aligned} \hbar- \|i_\ast F\|&> \|z_1 -z^{\prime}_1\|-\|z^{\prime}_1 -z_1^{+3}\|+ \|z_1 -z^{\prime}_2\|-\|z^{\prime}_2- z_1^{+3}\|\\[5pt] & = \sqrt{(\frac{k^2-g+1}{k})^2+4g} - \sqrt{(\frac{k^2-g+1}{k}-3)^2+4g} \\[5pt] & \mathrel{\phantom{>}} + \sqrt{(\frac{\tilde k^2-g+1}{\tilde k})^2+4g} - \sqrt{(\frac{\tilde k^2-g+1}{\tilde k}-3)^2+4g}\\[5pt] & \geq \frac{4k^2}{\sqrt{(g-1+k^2)^2+4k^2}+(g-1+k^2)} +\frac{4\tilde k^2}{\sqrt{(g-1+\tilde k^2)^2+4\tilde k^2}+(g-1+\tilde k^2)} \\[5pt] &=\hbar+\chi-2(g-1+k^2) \end{aligned} \end{align} $$

whenever $(g,k,m)\notin \Big \{(5,3,m), (6,4,3), (8,5,2)\Big \}$ satisfies our assumption.

In the case $(g, k,m)=(6,4,3)$ , $(8,5,2)$ or $(5,3,m)$ with $m\geq 4$ , though the inequality (6.3) fails, one can give an improvement of the estimate (6.3) by considering the convex hull of integer points in ${\mathbf {P}}_{0z_1'z_1^{+3}z_2'z_2}$ . In those cases, the convex hull is a convex polygon with vertices $z_1, z_1^{+3}, z_1', z_2'$ and $z_3$ , where $z_3$ is given as below:

• • $(g, k,m)=(5,3,m)$ , $z_3=-3+2\sqrt {-1}$ ;

• • $(g, k,m)=(6,4,3)$ , $z_3=-8+3\sqrt {-1}$ ;

• • $(g, k,m)=(8,5,2)$ , $z_3=-14+4\sqrt {-1}$ .

Then one can get

(6.4) $$ \begin{align} \hbar- \|i_\ast F\|> \|z_1\|-\|z_3\|- \|z_3-z_1^{+3}\|+ \|z_1 -z^{\prime}_2\|-\|z^{\prime}_2- z_1^{+3}\|. \end{align} $$

A computer calculation of their values shows that Equation (6.2) still holds.

In case (ii) and (iii), if $z_1^{+1}$ or $z_1^{+2}$ is a vertex of ${\mathbf {P}}_{i_*F}$ , there exists $\widetilde {E}_j \subset i_*F$ in the HN-filtration (5.11) such that $\overline Z(\widetilde {E}_j)=z_1^{+1}$ (respectively, $z_1^{+2}$ ). Then we have

$$ \begin{align*} \begin{aligned} h^0(X,i_*F) &\leq \sum_i \left\lfloor \frac{ l (E_i)+\chi(E_i)}{2}\right\rfloor \\[5pt] &\leq \sum_{i\leq j} (\left\lfloor \frac{l (E_i)+\chi(E_i)}{2}\right\rfloor -\frac{\chi(E_i)}{2}) + \sum_{i> j} (\left\lfloor \frac{l (E_i)+\chi(E_i)}{2}\right\rfloor-\frac{\chi(E_i)}{2} ) +\frac{\chi}{2}\\[5pt] & \leq \frac{\hbar+\chi}{2}. \end{aligned} \end{align*} $$

For simplicity, we may use $\frac {\hbar _1}{2}$ and $\frac {\hbar _2}{2}$ to denote the first two terms in the second row. As $h^0(X,i_\ast F)\geq g-1+k^2$ , following the argument in Theorem 5.6, it suffices to prove the inequality

$$ \begin{align*} \frac{\hbar + \chi}{2} - (g-1+k^2)< \frac{\hbar-\hbar_1-\hbar_2}{2}, \end{align*} $$

or equivalently,

(6.5) $$ \begin{align} \hbar_1+\hbar_2 < 2(g-1+k^2) -\chi=\|z_1^{+1}\| +\|z_1^{+1} -z_2\|-2. \end{align} $$

For case ii), a direct computation shows

(6.6) $$ \begin{align} \hspace{-120pt}\|z_1^{+1}\|-\hbar_1& = \sum_{i\leq j} \|E_i\|-\hbar_1-(\sum_{i\leq j}\|E_i\|-\|z_1^{+1}\|) \nonumber\\[5pt] & \geq \sum\limits_{i\leq j} (\|E_i\|- \ell(E_i))-(\sum_{i\leq j}\|E_i\|-\|z_1^{+1}\|) \nonumber \\[5pt] &\geq (\|E_1\|- \ell(E_1))-(\sum_{i\leq j}\|E_i\|-\|z_1^{+1}\|)\nonumber \\[5pt] & \geq \frac{3}{\sqrt{k^2+4g-3}}-(\sum_{i\leq j}\|E_i\|-\|z_1^{+1}\|) \end{align} $$

(6.7) $$ \begin{align} &\qquad\geq \frac{3}{\sqrt{k^2+4g-3}}-(\|z^{\prime}_1\| + \|z^{\prime}_1-z_1^{+1}\|-\| z_1^{+1}\| ) \end{align} $$

(6.8) $$ \begin{align} &>0.\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \end{align} $$

Let us explain why the inequality (6.6) holds. Note that $\overline {Z}(E_1)=x+y\sqrt {-1}$ satisfies

$$\begin{align*}k^2-g \leq -\frac{x}{y} \leq k^2-g+1.\end{align*}$$

Then we have $-x< ky$ and $y \nmid x$ by our assumption $g-1\neq k$ , $g\neq k$ , and $g<2k$ . This will give

$$ \begin{align*} \|E_1\|- \ell(E_1) &\geq \sqrt{x^2 +4g y^2}- \sqrt{x^2+4(g-1)y^2+y^2} \quad (\textrm{since } y\nmid x)\\ & \geq \frac{3y^2}{2 \sqrt{x^2+(4g-3)y^2}} \\ & \geq \frac{3y^2}{2 \sqrt{k^2y^2+(4g-3)y^2}} \\ &> \frac{3}{\sqrt{k^2+4g-3}} \quad (\textrm{since } y \geq 2, \text{ which is a consequence of } y\nmid x). \end{align*} $$

The inequality (6.8) holds because

$$ \begin{align*} \|z_1^{+1} -z^{\prime}_1 \|+\|z^{\prime}_1 \| -\|z_1^{+1}\|&= \sqrt{(\frac{k^2-(g-1)}{k}-1)^2+4g} + (k-1)\sqrt{(\frac{k^2-(g-1)}{k})^2+4g } -(k^2 +g) \\ &=\sqrt{(\frac{k^2-(g-1)}{k}-1)^2+4g}- \left(k-1+\frac{g+1}{k}\right)\\ &\mathrel{\phantom{=}}+(k-1)\sqrt{(\frac{k^2-(g-1)}{k})^2+4g } -\left((k^2 +g)-(k-1+\frac{g+1}{k})\right)\\ &< \frac{2g(k-1)}{k^3-k^2+(g+1)k} + \frac{-2 g (k-1)^2}{k^4-k^3+(g+1)k^2-(g+1)k}\\ &= \frac{2g(k-1)}{(k^2+g+1)(k^2-k+g+1)} \\ & \leq \frac{3}{\sqrt{k^2+4g-3}} \end{align*} $$

when $k\geq \frac {g+1}{2}\geq 2$ . Note that $ \|z_1^{+1}\|-\hbar _1=k^2+g- 2\sum \limits _{i\leq j} \left \lfloor \frac {l (E_i)+\chi (E_i)}{2}\right \rfloor -\sum \limits _{i\leq j} \chi (E_i)$ is an even number, this yields $ \|z_1^{+1}\|-\hbar _1 \geq 2.$

Next, recall that $\widetilde {E}_l=i_\ast F$ in the HN filtration (5.11), we can get

$$ \begin{align*} \|z_1^{+1} -z_2\|-\hbar_2 & \geq \sum\limits_{i>j} (\|E_i\|- \ell(E_i))-(\sum_{i> j}\|E_i\|-\|z_1^{+1}- z_2\|) \\ &\geq (\|E_l\|- \ell(E_l))-(\|z_1^{+1}-z^{\prime}_2\| + \|z^{\prime}_2- z_2\|-\|z_1^{+1} - z_2\| ) \\ &> \frac{3}{\sqrt{\tilde k^2+4g-3}} - \frac{2g(\tilde k-1)}{(\tilde k^2+g+1)(\tilde k^2-\tilde k+g+1)}>0 \end{align*} $$

as $\tilde {k}>\frac {g+1}{2}$ . Combining them together, we can obtain (6.5).

For case (iii), if $k\nmid g+1$ , we have

(6.9) $$ \begin{align} \|z_1^{+1}\|-\hbar_1& = \sum_{i\leq j}\|E_i\|-\hbar_1+\|z_1^{+1}\|-\sum_{i\leq j}\|E_i\| \nonumber \\ &\geq \|E_1\|-\ell(E_1) +\|z_1^{+1}\|-\|z^{\prime}_1\|-\|z^{\prime}_1-z_1^{+2}\| \nonumber \\ &\geq \frac{3}{\sqrt{k^2+4g-3}} + \|z_1^{+1}\|-\|z^{\prime}_1\|-\|z^{\prime}_1-z_1^{+2}\| \end{align} $$

(6.10) $$ \begin{align} &\kern1.2pt>1.\qquad\qquad\qquad\quad\qquad\qquad\qquad \end{align} $$

Here, the inequality (6.10) holds because

$$ \begin{align*} \|z_1^{+2}-z^{\prime}_1 \|+\|z^{\prime}_1\| -\|z_1^{+1}\| &=\sqrt{(\frac{k^2-(g-1)}{k}-2)^2+4g}+(k-1)\sqrt{(\frac{k^2-(g-1)}{k})^2+4g} -(k^2+g)\\ &=\sqrt{(\frac{k^2-(g-1)}{k}-2)^2+4g}-(k-2+\frac{g+1}{k})\\ &\mathrel{\phantom{\geq}} + (k-1)\sqrt{(\frac{k^2-(g-1)}{k})^2+4g}-(k^2-k+(g+1)-\frac{g+1}{k})-1 \\ &\leq \frac{2 g (2 k-1)}{k \left(g+(k-1)^2\right)}-\frac{2 g (k-1)}{k \left(g+k^2+1\right)}-1\\ &= \frac{2 g \left(k^2+2k+g-1\right)}{\left(k^2+g+1\right)\left(k^2-2k+g+1\right)} -1\\ &< \frac{3}{\sqrt{k^2+4g-3}}-1. \end{align*} $$

Note that $\|z_1^{+1}\|-\hbar _1$ is an odd number, this yields $\|z_1^{+1}\|-\hbar _1 \geq 3$ . Similarly, one can get

$$ \begin{align*} \|z_1^{+1}- z_2\|-\hbar_2\geq 3 \end{align*} $$

under the assumption $m(g-1)-k \nmid g+1 $ . Since both of them are at least positive under our assumption, we get Equation (6.5) as well. This finishes the proof for $(g,k,m)\neq (5,3,3)$ .

For the remaining case $(g,k,m)=(5,3,3)$ , we have to make use of the $4$ -sharpness of ${\mathbf {P}}_v$ . We just need to verify ${\mathbf {P}}_{i_*F}$ is not contained in ${\mathbf {P}}_{0z^{\prime }_1z_1^{+4} z_2' z_2}$ and $z_1^{+3}=-2+3\sqrt {-1}$ is not a vertex of ${\mathbf {P}}_{i_\ast F}$ . As above, by using the convex hull of integer points in ${\mathbf {P}}_{0z^{\prime }_1z_1^{+4} z_2' z_2}$ , we have

$$ \begin{align*} \begin{aligned} \hbar- \|i_\ast F\| \geq & \|z_1\|-\|-3+2\sqrt{-1}\|- \|-3+2\sqrt{-1}-z_1^{+4}\| + \|z_1 -z^{\prime}_2\|-\|z^{\prime}_2- z_1^{+4}\|\\ &=-2 \sqrt{6}-\sqrt{89}+\sqrt{205}+\frac{\sqrt{7549}}{9}-\frac{\sqrt{3301}}{9} \\ &>\hbar+\chi-2(g-1+k^2) \end{aligned} \end{align*} $$

which show that ${\mathbf {P}}_{i_*F}$ cannot lie in ${\mathbf {P}}_{0z^{\prime }_1z_1^{+4} z_2' z_2}$ . Moreover, a similar estimate of ${\|E_1\|-\ell (E_1)}$ and $\|E_l\|-\ell (E_l)$ in (ii) and (iii) shows that $z_1^{+3}$ is not a vertex of ${\mathbf {P}}_{i_\ast F}$ .

Proof. Let us check that the conditions (ii)–(iv) in Theorem 7.1 are satisfied. By our assumption, we know that $\gcd (r-s,c)=1$ . According to [Reference Feyzbakhsh13, Lemma 3.1], one has

$$ \begin{align*}h^0(X,E) \leq r+s,\end{align*} $$

which forces $h^0(X,E)= r+s $ by Equation (4.2). This verifies the condition (ii).

For the condition (iv), note that we have $s=\frac {c^2(g-1)}{r}\geq c^2$ where the equality holds when $r= g-1$ . If $r=g-1$ , the inequality in Equation (7.3) will be equality. By [Reference Feyzbakhsh11, lemma 3.1], we have $c\mid (g-1)$ which is a contradiction. Thus, we only need to verify the condition (iii). By Remark 3.5, it suffices to show

$$ \begin{align*}{\mathbf{P}}_{o \pi_{v_K} \pi_{v(-m)}} \setminus \{\text{vertices}\}\subseteq V(X).\end{align*} $$

To make the computation easier, we may consider the action of tensoring the invertible sheaf ${\mathcal {O}}_X(H)$ which sends the triangle ${\mathbf {P}}_{o \pi _{v_K} \pi _{v(-m)}}$ to the triangle ${\mathbf {P}}_{op_1p_2}$ , where $p_1=\pi _{v_K(1)}$ and $p_2=\pi _{v(1-m)}$ . Then it is equivalent to show there are no projection of roots in ${\mathbf {P}}_{op_1p_2}-\{\text {vertices}\}$ .

Firstly, we show that there is no projection of root on the two edges joining o. By definition, we have

$$ \begin{align*}p_1=(\frac{kc}{(k-1)^2 r},\frac{c}{(k-1)r}) \text{ and } p_2=(\frac{cr}{k((m-1)r-c)^2}, \frac{c}{k(c-(m-1)r)}). \end{align*} $$

Then two open edges ${\mathrm {L}}_{(o,p_1)}$ and ${\mathrm {L}}_{(o,p_2)}$ do not contain any projection of roots by Observation (B).

Next, since $(m-1)r>\frac {ck}{(k-1)}>c$ , we know that $p_1$ is lying in the first quadrant while projection $p_2$ is lying in the second quadrant. So the region

$$ \begin{align*}{\mathbf{P}}_{op_1p_2} \setminus ({\mathrm{L}}_{[o,p_1]}\cup {\mathrm{L}}_{[o,p_2]})\end{align*} $$

is contained in the union of two trapezoidal regions ${\mathbf {P}}_{o p_1 \infty }^\circ $ , ${\mathbf {P}}_{o p_2\infty }^\circ $ and the x-axis. As $r|(g-1)$ and $\gcd (r,c)=1$ , we have an inclusion

$$ \begin{align*}{\mathbf{P}}_{o p_2 \infty}^\circ\subseteq V(X)\end{align*} $$

from Lemma 4.2 (i). Moreover, if there is a root $\delta =(r',c',s')\in \mathrm {R}(X)$ with $r'>0$ whose projection $\pi _\delta $ is lying in ${\mathbf {P}}_{o p_1 \infty }^\circ $ , one can follow the computation in Lemma 4.2 to get inequalities

(7.4) $$ \begin{align} k c' < (k-1) r' < k c' +\frac{k}{(g-1)c'}. \end{align} $$

However, one can directly check that there are no such integers $(r',c',s')$ satisfying Equation (7.4) under the assumption $k\leq g-1$ or $3<k\leq 3g-3$ .

It remains to show that ${\mathbf {P}}_{op_1p_2} \cap y\text {-axis}\subseteq V(X)$ . Note that

$$ \begin{align*}{\mathrm{L}}_{[p_1, p_2]} \cap x\text{-axis}=( \frac{c/(k-1)}{(m-1)r-c},0)\end{align*} $$

which is below $o'$ . It follows that ${\mathbf {P}}_{op_1p_2} \cap y\text {-axis}\subseteq {\mathrm {L}}_{(o, o')}\subseteq V(X)$ .

Proof. By Corollary 5.7, $\psi $ is a bijective morphism. We only need to verify the conditions (ii)–(iv) in Theorem 7.1. We will check them one by one.

1. (1) Let us first verify that $h^0(X,E) =r+s$ for any $E\in \mathbf {M}(v)$ . By [Reference Feyzbakhsh13, Proposition 3.1], it suffices to show that

   (7.5) $$ \begin{align} \frac{\sqrt{(r-s)^2+ (2g+2)c^2}}{2} < \frac{r+s}{2} +1. \end{align} $$

   After simplification, one can find that Equation (7.5) is equivalent to

   $$\begin{align*}\frac{(g+1)c^2-1}{2r+2}-1 <s. \end{align*}$$

   This holds when $(g-1)c^2 -rs <r$ and $g> 4r^2+1$ .

2. (2) For condition (ii), we claim that

   $$ \begin{align*}{\mathrm{L}}_{(\pi_{v(-m)},\pi_{v_K})}\cap \Gamma \neq \emptyset,\end{align*} $$
   and hence ${\mathrm {L}}_{(\pi _{v(-m)},\pi _{v_K})}\cap V(X)\neq \emptyset $ . Let us write $v(-m)=(r, \tilde {c},\tilde {s} )$ and $v_K=(s,-c,r)$ with $\tilde {c}=c-mr$ and $ \tilde s=\lfloor \frac {(g-1)\tilde c^2 +1}{r}\rfloor ~ $ . Then we only need to show that the quadratic equation
   (7.6) $$ \begin{align} g((1-t)t\tilde{c}+t(-c))^2 = ((1-t)r+ t s)((1-t)\tilde{s}+ t r) \end{align} $$
   has roots for $0<t<1$ . By calculating the discriminant of (7.6), we know it has a solution $t_0$ satisfying
   (7.7) $$ \begin{align} 0<t_0 < \frac{\tilde{c}^2 g -r\tilde{s}}{s\tilde{s}+r^2 +2\tilde c cg+2(\tilde c^2 g-r\tilde s)}<1. \end{align} $$

3. (3) Choose $\sigma \in {\mathrm {L}}_{(\pi _{v(-m)}, \pi _{v_K})} \cap \Gamma $ . We first verify that

   $$ \begin{align*}\Omega^+_{v(-m)}({\mathrm{L}}_{ (o,\sigma]}) \cap {\mathrm{H}}_{\text{alg}}^{*}(X)= \emptyset.\end{align*} $$

   Suppose there is an integer point $p_0=(x_0,y_0,z_0) \in \Omega ^+_{v(-m)}({\mathrm {L}}_{[ \sigma , o)})$ . By Lemma 7.4 below, we have

   $$ \begin{align*}\tilde{c}-1 < y_0< 0.\end{align*} $$

   Moreover, one may observe that $p_0$ is lying in a (closed) planer region enclosed by the conic

   $$ \begin{align*} Q= \left\{ y = y_0 ,\; (g-1)y^2+1= xz \right\} \end{align*} $$
   and two lines
   $$ \begin{align*}L_1=\left\{y = y_0,\; z=\frac{y_0 \tilde s}{\tilde{c}} \right\};~ L_2 =\left\{ (1-t)\frac{y_0}{\tilde{c}}(r,\tilde{c},\tilde{s} )-\frac{ty_0}{c}(s,-c,r) ,t\in \mathbb{R} \right\}.\end{align*} $$

   It has three vertices given by the intersection points $L_1\cap L_2$ , $L_1\cap Q$ and $L_2\cap Q$ . This yields

   (7.8) $$ \begin{align} \begin{aligned} \frac{y_0 r}{\tilde c} \leq x_0 \leq (1-t') \frac{y_0r}{\tilde c} -\frac{t'y_0 s}{c}, \end{aligned} \end{align} $$
   where $t'$ is the smaller root of the quadratic equation
   $$ \begin{align*}(g-1)y_0^2+1= [(1-t) \frac{y_0r}{\tilde c} -\frac{ty_0 s}{c} ] [(1-t) \frac{y_0\tilde s}{\tilde c} -\frac{ty_0 r}{c}].\end{align*} $$

   Solving the equation, one can get

   (7.9) $$ \begin{align} \begin{aligned} t' \leq 2 \frac{(g-1) y_0^2+1-\frac{r\tilde s y_0^2}{\tilde c^2}}{-\frac{\tilde c \tilde s s}{c} -2r\tilde s} \leq \frac{2c(y_0^2r+\tilde c^2)}{ (-\tilde c\tilde s s-2r c\tilde s)\tilde c^2} \end{aligned} \end{align} $$
   as $(g-1)\tilde c^2-r\tilde s< r$ . Plugging Equation (7.9) into (7.8), we get
   $$ \begin{align*} \begin{aligned} 0< x_0 -\frac{y_0r}{\tilde c}&\leq \frac{2cy_0(y_0^2r+\tilde c^2)}{ (\tilde c\tilde s s+2r c\tilde s)\tilde c^2} (\frac{s\tilde c+rc}{c\tilde c}) \\&= (\frac{y_0}{\tilde c})(\frac{y_0^2r+\tilde c^2}{\tilde s\tilde c^2}) (\frac{s\tilde c+rc}{s\tilde c+2rc}) \\ &\leq \frac{3(r+1)}{\tilde s } \\ & < \frac{3r(r+1)}{(g-1)\tilde{c}^2-r} \\ &< -\frac{1}{\tilde c}, \end{aligned} \end{align*} $$
   where the last inequality holds because $g-1\geq 4r^2$ . This contradicts to $x_0,y_0\in \mathbb {Z}$ . A similar computation shows that $\Omega ^+_{v_K}({\mathrm {L}}_{(o, \sigma ]}) \cap {\mathrm {H}}_{\text {alg}}^{*}(X)=\emptyset $ as well.

4. (4) Condition (iv) holds since our assumption $g>4r^2+1 $ ensures that

   $$\begin{align*}2s> 2r-2 +2c^2 >v^2 + 2c^2.\\[-37pt] \end{align*}$$

Proof. Set $v_{t} = (1-t) v(-m) + t v_K$ . Let $0<t_0<t_1<1 $ be the roots of Equation (7.6). We set

$$ \begin{align*}w =(x',y',z')= {\mathrm{L}}_{O,v_{t_0}} \cap {\mathrm{L}}_{v(-m), v(-m) + v_{t_1}}.\end{align*} $$

Then $\Omega ^+_{v(-m)}({\mathrm {L}}_{(o,\sigma ]})$ is contained in the tetrahedron $ \mathbf {T}_{Ov(-m)\varpi w}$ with four vertices $O, v(-m), w $ and $\varpi = (r, \tilde {c}, \frac {g\tilde {c}^2}{r})$ . This gives $y'<y_0<0 $ . Hence, we only need to estimate the lower bound of $y'$ .

Set $v_{t_0}=(r_{t_0},c_{t_0}, s_{t_0})$ , then we have $w=\frac {y'}{c_{t_0}}v_{t_0}\in {\mathrm {L}}_{o,v_{t_0}}$ . Note that $w - v(-m)\in {\mathrm {L}}_{O, \, v_{t_1}}$ is lying on the hyperboloid

$$ \begin{align*}\Big\{(x,y,z)\in \mathbb{R}^3|~g y^2 -x z =0\Big\}.\end{align*} $$

Then one can see that $y'<\tilde c-1$ if

(7.10) $$ \begin{align} \begin{aligned} &\frac{\tilde{c}-1}{c_{t_0}} v_{t_0} -v(-m) \in \Big\{(x,y,z)\in \mathbb{R}^3|~g y^2 -x z <0\Big\}, \end{aligned} \end{align} $$

that is,

(7.11) $$ \begin{align} \big( \frac{\tilde{c}-1}{c_{t_0} }r_{t_0} -r\big) \big(\frac{\tilde{c}-1}{c_{t_0} }s_{t_0}-\tilde{s} \big)>g. \end{align} $$

Plugging the coordinates of $v_{t_0}$ into Equation (7.11) and simplify all the terms, one can obtain a quadratic inequality of $t_0$ and one can easily see that Equation (7.11) holds if

$$\begin{align*}t_0 < \frac{(2-\tilde{c})(\tilde{c}^2 g -r\tilde{s})}{ \tilde{c}[r\tilde{s}-r^2-s\tilde{s}+(2-\tilde{c})g(c+\tilde{c}) ] + r^2+s\tilde{s} -(c+2) r\tilde{s}}. \end{align*}$$

Using the upper bound of $t_0$ given in Equation (7.7), we are reduced to check

$$\begin{align*}\begin{aligned} \tilde{c}(r\tilde{s}-r^2-s\tilde{s}+(2-\tilde{c})g(c+\tilde{c}) ) + r^2+s\tilde{s} -(c+2) r\tilde{s} < (2-\tilde{c})(s\tilde{s}+r^2-2r\tilde{s}+2\tilde{c}g(\tilde{c}+c)). \end{aligned} \end{align*}$$

After further simplification and reduction, the inequality above becomes

(7.12) $$ \begin{align} \begin{aligned} 0 <s\tilde s+r^2+2\tilde c g+c-(\tilde c+c-2)(\tilde c^2 g-r\tilde s). \end{aligned} \end{align} $$

The right-hand side can be estimated as below

(7.13) $$ \begin{align} \begin{aligned} \mathrm{RHS}&> r^2+ s\tilde s+2\tilde c g-(\tilde c+c-2)(\tilde c^2 g-r\tilde s)\\ &= r^2+ s\tilde s +2\tilde c g+(mr+2-2c) (\tilde c^2+\frac{v^2}{2})\\ & > r^2+s\tilde s +2\tilde c g-r (\tilde c^2+r)\\ &\geq (\frac{(g-1)c^2}{r}-1)(\frac{(g-1)\tilde{c}^2}{r}-1) +2\tilde c g-r\tilde c^2 \\ &>0.\\[-16pt] \end{aligned} \end{align} $$

Proof. Note that v satisfies the injectivity condition (✶), $\psi $ is an injective morphism with stable image. It also satisfies the condition (✶✶✶). Due to Theorem 7.2, it suffices to show that $\psi $ is surjective. The idea is to use Theorem 5.3. Suppose one has

$$ \begin{align*}{\mathbf{P}}_v\neq {\mathbf{P}}_{i_\ast F}\end{align*} $$

for some $F\in {\mathbf {BN}}_C(v)$ . A direct computation shows ${\mathbf {P}}_v$ is at least $2$ -sharp. Then either ${\mathbf {P}}_{i_*F}$ lies inside the polygon ${\mathbf {P}}_{0 z_1^{+2} z^{\prime }_2 z_2}$ , or it has $z_1^{+1}$ as a vertex. For the first case, one has

$$ \begin{align*} \hbar- \|i_\ast F\| &> \|z_1- z^{\prime}_1\|-\|z^{\prime}_1- z_1^{+2}\|+ \|z_1- z^{\prime}_2\|-\|z^{\prime}_2- z_1^{+2}\| \\ &=\frac{\sqrt{877}}{3}-\frac{\sqrt{613}}{3} \\&> \sqrt{29}+\sqrt{877}-34 ={\hbar+ \chi} - 10 \end{align*} $$

which contradicts Equation (5.18). For the second case, it forces $\overline Z(\widetilde {E}_1)=z_1^{+1}$ and hence

$$ \begin{align*} 5 \leq h^0(C, F)= h^0(X,i_*F) \leq \lfloor \frac{\ell( z_1^{+1})}{2} \rfloor+ \frac{\|z_1^{+1}- z^{\prime}_2\|+\|z^{\prime}_2 -z_2\|+\chi}{2} \leq \frac{\hbar+\chi}{2}. \end{align*} $$

However, we have

$$ \begin{align*} \begin{aligned} \frac{\hbar-(\|z_1^{+1} -z^{\prime}_2\|+\|z^{\prime}_2- z_2\|)}{2}-\lfloor \frac{\ell( z_1^{+1})}{2} \rfloor&= \frac{1}{6} \left(\sqrt{877}-4 \sqrt{46}\right)+\frac{\sqrt{7}}{2}-1 \\&> \frac{1}{2} \left(\sqrt{29}+\sqrt{877}-34\right)\\&= \frac{\hbar+\chi}{2}-5 \end{aligned} \end{align*} $$

which is impossible. It follows from Theorem 5.3 that $\psi $ is surjective.

Proof. The idea is to use the bound for prime character nonresidues. In [Reference Pollack19, Theorem 1.3], it has been proved that there exists an integer $m_0$ with the property: if $j> j_0$ and $\chi $ is a quadratic character modulo j, there are at least $\log (j)$ primes $\ell \leq \sqrt [3]{j}$ with $\chi (\ell )= 1$ . Choose N to be the minimal integer satisfying

• • $8(N-1)n\geq j_0$ ,

• • the $\lfloor \log (8(N-1)n)\rfloor $ -th prime number $>n+1$ ,

• • $\sqrt [3]{8(N-1)n}\leq \frac {\sqrt {N-1}}{2}$ .

Clearly, N only depends on n. For $g>N$ , we write

$$ \begin{align*}(g-1)n=a^2\prod_{i=1}^k q_i ,\end{align*} $$

where $q_i$ are distinct primes. Let $\chi _i$ be the character defined by

$$ \begin{align*}\chi_i(d)= \left( \frac{d }{q_i}\right)(-1)^{\frac{(d-1)(q_i-1)}{4}}\end{align*} $$

if $q_i$ is odd and $\chi _i(d)=(-1)^{\frac {d^2-1}{8}}$ if $q_i=2$ . Consider the quadratic character

$$ \begin{align*}\chi(d)=\prod_{i=1}^k \chi_i(d)\end{align*} $$

modulo $8(g-1)n$ . As $8(g-1)n>N\geq j_0$ , there exists a prime p such that $\chi (p)=1$ and

$$ \begin{align*}n+1<p<\sqrt[3]{8(g-1)n}\leq\frac{ \sqrt{g-1}}{2}.\end{align*} $$

Moreover, one can compute the Jacobi symbol

$$ \begin{align*}\left( \frac{n(g-1)}{p}\right)=\prod_{i=1}^k \left( \frac{q_i }{p}\right)=\prod_{i=1}^k \chi_i(p) =\chi(p)=1\end{align*} $$

by the law of reciprocity. It follows that $x^2=(g-1)n\ \mod p$ has a solution.